Consider the following two equivalent definitions of an open set in a metric space $(X,d)$:

  • Definition 1: A set $U$ in a metric space is open if and only if $U$ is an arbitrary union of open balls of elements in the metric space, or an open ball itself.

  • Definition 2: A set $U$ in a metric space is open if and only if, for every element $x\in X$, there exists an $\epsilon$ such that an open ball $B_{\epsilon}(x)$ is contained within $U$.

Although Definition 2 clearly implies Definition 1, how does Definition 1 imply Definition 2?

  • $\begingroup$ For a set $U$ described in definition 2, it can be represented as the union of all of the open ball witnin $U$ itself containing some $x \in U$ , and the index can run over all of $U$, which means every $x\in U$ belongs to one of the such ball. $\endgroup$
    – Egyptian
    Commented Mar 23, 2022 at 12:42

1 Answer 1


Definition 1 implies 2 because of the observation that if $B(x,r)$ is an open ball and $y\in B(x,r)$ then for $r'=r-d(x,y)$ (which is positive) we have $B(y,r')\subseteq B(x,r)$. The last inclusion follows from the triangle inequality.


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